Monday, Aug. 2, 2004PHYS 1441-501, Summer 2004 Dr. Jaehoon Yu 1 PHYS 1441 – Section 501 Lecture #17 Monday, Aug. 2, 2004 Dr. Jaehoon Yu Flow Rate and Equation

Monday, Aug. 2, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu

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PHYS 1441 – Section 501Lecture #17

Monday, Aug. 2, 2004Dr. Jaehoon Yu

• Flow Rate and Equation of Continuity• Bernoulli’s Equation• Simple Harmonic Motion• Simple Block Spring System• Energy of a Simple Harmonic Oscillator

Final term exam next Wednesday, Aug. 11!


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Announcements

• Final term exam Wednesday, Aug. 11– Time: 6:00 – 8:00 pm– Location: SH125– Covers: Section 8.6 – Ch. 11– Review on Monday, Aug. 9– Mixture of multiple choice and free style

• Please do not miss the exam


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Flow Rate and the Equation of ContinuityStudy of fluid in motion: Fluid Dynamics

If the fluid is water: •Streamline or Laminar flow: Each particle of the fluid follows a smooth path, a streamline•Turbulent flow: Erratic, small, whirlpool-like circles called eddy current or eddies which absorbs a lot of energy

Two main types of flow

Water dynamics?? Hydro-dynamics

Flow rate: the mass of fluid that passes a given point per unit time /m t

since the total flow must be conserved

1m

t

1 1V

t

1 1 1A l

t

1 1 1Av

1 1 1Av 1m

t

Equation of Continuity

2m

t

2 2 2A v


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Example for Equation of ContinuityHow large must a heating duct be if air moving at 3.0m/s along it can replenish the air every 15 minutes, in a room of 300m3 volume? Assume the air’s density remains constant.

Using equation of continuity

1 1 1Av

Since the air density is constant

1 1Av Now let’s imagine the room as the large section of the duct

1A 2 2

1

/A l t

v 2

1

V

v t

2300

0.113.0 900

m

2 2 2A v

2 2A v

2 2

1

A v

v


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Bernoulli’s EquationBernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high.

Amount of work done by the force, F1, that exerts pressure, P1, at point 1

1W

Work done by the gravitational force to move the fluid mass, m, from y1 to y2 is

1 1F l 1 1 1PA lAmount of work done on the other section of the fluid is

2W

3W 2 1mg y y

2 2 2P A l


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Bernoulli’s Equation cont’dThe net work done on the fluid is

W 1 1 1PA l From the work-energy principle

22

1

2mv

Since mass, m, is contained in the volume that flowed in the motion

1 1A l and m

2 2221

2A l v

1 1 1PA l

1 11 lPA

Thus,

2 2 2P A l 2 1mgy mgy 1W 2W 3W

21

1

2mv 2 2 2P A l 2 1mgy mgy

2 2A l 1 1A l 2 2A l

1211

1

2vA l

2 22 lP A 2 2 1 12 1gyA l A l gy


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Bernoulli’s Equation cont’d

We obtain

2 2 1 1 1 1 2 2 2 2 12 22 1 1 2 2 11

1 1

2 2A l A l A l A l A lv v P P g A ly gy

Re-organize

21 1 1

1

2P v gy Bernoulli’s

Equation

21 1 1

1

2P v gy

22 2 2

1

2P v gy

2 22 1 1 2 2 1

1 1

2 2v v P P gy gy

Since

Thus, for any two points in the flow

For static fluid 2P

For the same heights 2 22 1 1 2

1

2P P v v

The pressure at the faster section of the fluid is smaller than slower section.

Pascal’s Law

.const

1 1 2P g y y 1P gh


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Example for Bernoulli’s EquationWater circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter pipe on the second 5.0m above? Assume the pipes do not divide into branches.

Using the equation of continuity, flow speed on the second floor is

2v

Using Bernoulli’s equation, the pressure in the pipe on the second floor is

2P

5 3 2 2 313.0 10 1 10 0.5 1.2 1 10 9.8 5

2

5 22.5 10 /N m

1 1

2

Av

A

21 122

r v

r

2

0.0200.5 1.2 /

0.013m s

1P 2 21 2

1

2v v 1 2g y y


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Vibration or OscillationWhat are the things that vibrate/oscillate?

A periodic motion that repeats over the same path.

A simplest case is a block attached at the end of a coil spring.

Acceleration is proportional to displacement from the equilibriumAcceleration is opposite direction to displacement

• Tuning fork• A pendulum• A car going over a bump• Building and bridges• The spider web with a prey

So what is a vibration or oscillation?

kxFThe sign is negative, because the force resists against the change of length, directed toward the equilibrium position.

When a spring is stretched from its equilibrium position by a length x, the force acting on the mass is

This system is doing a simple harmonic motion (SHM).


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Vibration or Oscillation PropertiesThe maximum displacement from the equilibrium is

The complete to-and-fro motion from an initial point

One cycle of the oscillation

Amplitude

Period of the motion, TThe time it takes to complete one full cycle

Frequency of the motion, f

The number of complete cycles per second

Unit?

s-1Unit?

s

Relationship between period and frequency?

fT

1 T

1

for


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Vibration or Oscillation Properties

Amplitude? A• When is the force greatest?• When is the velocity greatest?• When is the acceleration greatest?• When is the potential energy greatest?• When is the kinetic energy greatest?


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Example 11-1

(a)

Car springs. When a family of four people with a total mass of 200kg step into their 1200kg car, the car’s springs compress 3.0cm. (a) What is the spring constant of the car’s spring, assuming they act as a single spring? (b) How far will the car lower if loaded with 300kg?

200 9.8 1960F mg N What is the force on the spring?

From Hooke’s law

(b)

F kx

Now that we know the spring constant, we can solve for x in the force equation

F kx

k

0.03k 1960mg N gF

x mg

x 41960

6.5 10 /0.03

N m

x

mg 300 9.8

mg

k 2

4

300 9.84.5 10

6.5 10m


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Energy of the Simple Harmonic OscillatorHow do you think the mechanical energy of the harmonic oscillator look without friction?Kinetic energy of a harmonic oscillator isThe elastic potential energy stored in the spring

Therefore the total mechanical energy of the harmonic oscillator is

KEPE

E

Total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude.

2

2

1mv

2

2

1kx

PEKE 2 21 1

2 2mv kx


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Energy of the Simple Harmonic Oscillator cont’dMaximum KE is when PE=0 maxKE

The speed at any given point of the oscillation

E

2max2

1mv 2

2

1 k

PEKE 22

2

1

2

1kxmv 2

2

1 k

v 22 xAmk 2

max 1x

vA

xA-A

KE/PEE=KE+PE=kA2/2

Maximum speed maxv kA

m


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Example 11-3Spring calculations. A spring stretches 0.150m when a 0.300-kg mass is hung from it. The spring is then stretched an additional 0.100m from this equilibrium position and released.

From Hooke’s law

0.100A m

0.300 9.8F kx mg N

(a) Determine the spring constant.

(b) Determine the amplitude of the oscillation.

Since the spring was stretched 0.100m from equilibrium, and is given no initial speed, the amplitude is the same as the additional stretch.

0.300 9.819.6 /

0.150

mgk N m

x


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Example cont’d

Maximum acceleration is at the point where the mass is stopped to return.

a

(c) Determine the maximum velocity vmax.

maxv kA

m

19.60.100 0.808 /

0.300m s

(d) Determine the magnitude of velocity, v, when the mass is 0.050m from equilibrium.

v2

max 1x

vA

20.050

0.808 1 0.700 /0.100

m s

(d) Determine the magnitude of the maximum acceleration of the mass.

kA

m 219.6 0.100

6.53 /0.300

m s

F ma kA Solve for a


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Example for Energy of Simple Harmonic OscillatorA 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00 cm.

The total energy of the cube is E

From the problem statement, A and k are mNk /0.20

Maximum speed occurs when kinetic energy is the same as the total energy

2maxmax 2

1mvKE

mcmA 03.000.3

PEKE 2

2

1kA J32 1000.903.00.20

2

1

E 2

2

1kA

maxvm

kA sm /190.0

500.0

0.2003.0


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Example for Energy of Simple Harmonic Oscillatorb) What is the velocity of the cube when the displacement is 2.00 cm.

velocity at any given displacement is

v

c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.

Kinetic energy, KE

KE

Potential energy, PE PE

22 xAmk

sm /141.0500.0/02.003.00.20 22

2

2

1mv J32 1097.4141.0500.0

2

1

2

2

1kx J32 1000.402.00.20

2

1


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Sinusoidal Behavior of SHM


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The Period and Sinusoidal Nature of SHM

v0

Consider an object moving on a circle with a constant angular speed

If you look at it from the side, it looks as though it is doing a SHM

sin2

0 1x

v vA

0v0

2 AT

v

20

1

2mv

2m

Tk

1 1

2

kfT m

Since it takes T to complete one full circular motion

From an energy relationship in a spring SHM

0

kv A

m

Thus, T is

Natural Frequency

0

v

v

2 2A x

A

2

1x

A

2 A

T

2 Af

21

2kA


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Example 11-5Car springs. When a family of four people with a total mass of 200kg step into their 1200kg car, the car’s springs compress 3.0cm. The spring constant of the spring is 6.5x104N/m. What is the frequency of the car after hitting the bump? Assume that the shock absorber is poor, so the car really oscillates up and down.

f

T 2m

k 4

14002 0.92

6.5 10s

1

T

1

2

k

m

41 6.5 101.09

2 1400Hz


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Example 11-6Spider Web. A small insect of mass 0.30 g is caught in a spider web of negligible mass. The web vibrates predominantly with a frequency of 15Hz. (a) Estimate the value of the spring constant k for the web.

f

(b) At what frequency would you expect the web to vibrate if an insect of mass 0.10g were trapped?

f

k

Solve for k1

2

k

m 15Hz

2 24 mf 22 44 3 10 15 2.7 /N m

1

2

k

m

4

1 2.726

2 1 10Hz


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The SHM Equation of Motion

v0

The object is moving on a circle with a constant angular speed

How is x, its position at any given time expressed with the known quantities?

cosx A t cosx A t

since

cos 2A ftHow about its velocity v at any given time?

2 f and

v 0 sinv 0 sinv t 0 sin 2v ftHow about its acceleration a at any given time?

0

kv A

m

a F

m

kx

m

cos 2

kAft

m

0 cos 2a ftFrom Newton’s 2nd law

0

kAa

m


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Sinusoidal Behavior of SHM

cos 2x A ft 0 sin 2v v ft

0 cos 2a a ft


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The Simple PendulumA simple pendulum also performs periodic motion.

The net force exerted on the bob is

rF

x L

Satisfies conditions for simple harmonic motion!It’s almost like Hooke’s law with.

Since the arc length, x, is

tF F mg

kL

The period for this motion is T

The period only depends on the length of the string and the gravitational acceleration

AmgT cos 0

tF Amg sin mg

mgx

L

2m

k 2

mL

mg 2

L

g


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Example 11-8Grandfather clock. (a) Estimate the length of the pendulum in a grandfather clock that ticks once per second.

Since the period of a simple pendulum motion is T

The length of the pendulum in terms of T is 2

2

4gT

L

Thus the length of the pendulum when T=1s is

2

24

T gL

g

L2

(b) What would be the period of the clock with a 1m long pendulum?

g

L2T1.0

2 2.09.8

s

2

1 9.80.25

4m


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Damped OscillationMore realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.

How do you think the motion would look?

Amplitude gets smaller as time goes on since its energy is spent.

Types of dampingA: Overdamped

B: Critically damped

C: Underdamped


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Forced Oscillation; ResonanceWhen a vibrating system is set into motion, it oscillates with its natural frequency f0.

However a system may have an external force applied to it that has its own particular frequency (f), causing forced vibration.

For a forced vibration, the amplitude of vibration is found to be dependent on the different between f and f0. and is maximum when f=f0.

A: light dampingB: Heavy damping

The amplitude can be large when f=f0, as long as damping is small.

This is called resonance. The natural frequency f0 is also called resonant frequency.

0

1

2

kf

m

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Monday, Aug. 2, 2004PHYS 1441-501, Summer 2004 Dr. Jaehoon Yu 1 PHYS 1441 – Section 501 Lecture #17 Monday, Aug. 2, 2004 Dr. Jaehoon Yu Flow Rate and Equation